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3 years ago

What is m KJL? ( thank you for anyone who answers!!)

Mathematics

1 answer:

jasenka [17]3 years ago

6 0

Answer:

54 degrees. The triangle is isosceles, so 180-72= 108.

108/2=54

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Naomi buys last year's best-selling novel, in hardcover, for $24.00. This is with a 25% discount from the original price. What w

RoseWind [281]

Turn 25% into a decimal by moving it twice to the left them adding ir decimal point. then multiply .25 ( what u get from 25%) to 24. once u get the product add ur decimal point
. U should get 6

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3 years ago

Does anyone know how tj do this (trig)

Crank

Answer:

x = 5

Step-by-step explanation:

NOTE: THE 2 SIDES ARE THE SAME LENGTH

THEY HAVE TO BE SMALLER THEN 10

So 10/2 = 5

x = 5

<em><u>Please mark as brainliest </u></em>

Have a great day, be safe and healthy

Thank u

8 0

3 years ago

Emerson purchased a game that was on sale for 18% off. The sales tax in his county is 8%. Let y represent the original price of

Juli2301 [7.4K]

Answer:

Step-by-step explanation:

Let us see...

y is the price.

Then, 18% off. This is:

y - (y*0.18) = 0.82*y

Then, we add 8%, so now it's 1.08% of this expression:

8 0

3 years ago

Read 2 more answers

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

GEOMETRY HELP AHH please

KATRIN_1 [288]

Answer:

Step-by-step explanation:

6 0

3 years ago